Problem: Find $\lim_{\theta\to -\scriptsize\dfrac{\pi}{4}}\dfrac{1+\sqrt 2\sin(\theta)}{\cos(2\theta)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $\dfrac{1}{2}$ (Choice C) C $\dfrac{1}{4}$ (Choice D) D The limit doesn't exist
Answer: Substituting $\theta=-\dfrac{\pi}{4}$ into $\dfrac{1+\sqrt 2\sin(\theta)}{\cos(2\theta)}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since our expression includes trigonometric functions, let's try to re-write it using factorization and trigonometric identities. Since we have $\cos(2\theta)$ in our expression, let's rewrite it using one of its double angle identities. Since the expression in the numerator includes $\sin(\theta)$, the most fitting identity is $\cos(2\theta)=1-2\sin^2(\theta)$. $\begin{aligned} &\phantom{=}\dfrac{1+\sqrt 2\sin(\theta)}{\cos(2\theta)} \\\\ &=\dfrac{1+\sqrt 2\sin(\theta)}{1-2\sin^2(\theta)} \gray{\cos(2\theta)\text{ identity}} \\\\ &=\dfrac{1+\sqrt 2\sin(\theta)}{(1+\sqrt 2\sin(\theta))(1-\sqrt 2\sin(\theta))} \gray{\text{Diff. of squares}} \\\\ &=\dfrac{\cancel{1+\sqrt 2\sin(\theta)}}{\cancel{(1+\sqrt 2\sin(\theta))}(1-\sqrt 2\sin(\theta))} \gray{\text{Cancel common factors}} \\\\ &=\dfrac{1}{1-\sqrt 2\sin(\theta)}\text{, for }\theta\neq \{...,-\dfrac{3}{4}\pi, -\dfrac{1}{4}\pi, \dfrac{5}{4}\pi, \dfrac{7}{4}\pi,...\} \end{aligned}$ This means that the two expressions have the same value for all $\theta $ -values (in their domains) except for $-\dfrac{1}{4}\pi+2\pi k$ or $-\dfrac{3}{4}\pi+2\pi k$ for any integer $k$, and specifically $-\dfrac{\pi}{4}$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{1+\sqrt 2\sin(\theta)}{\cos(2\theta)}=\dfrac{1}{1-\sqrt 2\sin(\theta)}$ for all $\theta$ -values in the interval $(-1,0)$ except for $\theta=-\dfrac{\pi}{4}$. Therefore, $\lim_{\theta\to -\scriptsize\dfrac{\pi}{4}}\dfrac{1+\sqrt 2\sin(\theta)}{\cos(2\theta)}=\lim_{\theta\to -\scriptsize\dfrac{\pi}{4}}\dfrac{1}{1-\sqrt 2\sin(\theta)}=\dfrac{1}{2}$ (The last limit was found using direct substitution.) In conclusion, $\lim_{\theta\to -\scriptsize\dfrac{\pi}{4}}\dfrac{1+\sqrt 2\sin(\theta)}{\cos(2\theta)}=\dfrac{1}{2}$.